Question

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10\, m$. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point $C$ from point $A$ is $t(\sqrt{2}+1)$ s. The value of $t$ is. _____(use $g=10\, m / s ^2$ )

Answer 1

From E.C. $=\frac{1}{2} mv _0^2= mgh$
$v _0=10 \sqrt{2}$
For $A \rightarrow B$
at $B , v =0$
$a=-g \sin 45^{\circ}=\frac{-10}{\sqrt{2}}$
$v = u + at _1 \Rightarrow 0=10 \sqrt{2}-\frac{10}{\sqrt{2}} t _1 \Rightarrow t _1=2 sec$
For $B \rightarrow C$
$s = ut _2+\frac{1}{2} at _2^2 $
$ \frac{10}{\sin 30^{\circ}}=\frac{1}{2}\left(10 \sin 30^{\circ}\right) t _2^2 $
$t _2=2 \sqrt{2}$
So total time
$T = t _1+ t _2 $
$ =2 \sqrt{2}+2$
$=2(\sqrt{2}+1) sec$